# Triangular load on a two-point support beam

• Sep 27th 2012, 06:57 AM
Vincent4312
Triangular load on a two-point support beam
Attachment 24943

Ok this is going to be a little hard to explain since my "math english" isn't that good.
We know that the amount the two point support beam bends under the net force P of the triangular load is
Attachment 249440<=x<=L
where e and i are positive constants associated to the support beam's material and cross-section. In what point the bending on the support beam is largest when L=3 meters?
I hope you understand it :D. Thanks.
• Sep 27th 2012, 07:42 AM
johnsomeone
Re: Triangular load on a two-point support beam
You explained it pretty well. I'm not sure where you're having trouble, so I'll just give a few hints/suggestions:

1) I like to de-clutter such problems. I'd set u = x/L and k = PL^3/(180ei), then work the problem for f(u) = k(3u^5 - 10 u^3 + 7u).
Since 0<=x<=L, have 0<=u<=1. (In the end, of course, will have to convert back to x and use L = 3 m.)

2) Sometimes a 4th degree equation is a quadratic equation in a square power. i.e. z^4 - 6z^2 - 10 = 0 is solveable
It's (z^2)^2 - 6(z^2) - 10 = 0, so if w = z^2, it's w^2 - 6w - 10 = 0, so w = [-(-6) +- sqrt(36-(-40)]/2 = 3+- sqrt(19).
Then z = +- sqrt[ 3+ sqrt(19)] and z = +- sqrt[sqrt(19)-3] i

Can you say where you're having difficulty with this problem?
• Sep 27th 2012, 07:45 AM
ebaines
Re: Triangular load on a two-point support beam
You will have to clarify something: what do you mean by "where bending is greatest?" If you mean where the deflection of the beam is greatest that would be where the slope of the beam deflection = 0, so take the derivative of y with respect to position x, set it to 0, and solve for x. But if you mean where the curvature of the beam is greatest that would be where the 2nd derivative of y is at a max (which also happens to be where the moment in the beam is greatest, and where the shear is 0).

I assume this problem comes from a course you are taking in beam theory - a mechanical or civil engineering class, right? You may find it good practice work to develop the beam equation from fundamentals. To do this you would need to;

1. Find the reaction forces for the two suports, F1 and F2
2. Develop the equation for shear V(x) = integral of p(x), using the boundary condition that V(0) = F1 and V(L) = F2. To find where bending curvature is greatest, solve for v(x)=0.
3. Integrate V(x) to get the equation for the moment in the beam, M(x), using the boundary condition for simple supports that M(0)=M(L)=0.
4. Integrate the M(x) equation and divide by EI to get the equation for the slope of the deflection. We don't have any convenient boundary conditions to apply for this step, so don't forget to include the constant of integration.
5. Integrate the slope equation to get the deflection equation y(x), and using boundary conditions y(0)= y(L)= 0 you can solve for the unknown constant from step 4.

Like I said, it's good practice!
• Sep 27th 2012, 09:01 AM
bjhopper
Re: Triangular load on a two-point support beam
From Handbook American Inst of Steel Construction for load increasing uniformerly to one end

R1 low end of load = W/3 W in kips
R2 high end of load = 2W/3
Max moment @ X from R1 = l (span in inches )/rad3 = 2Wl/9rad3
• Sep 27th 2012, 01:21 PM
Vincent4312
Re: Triangular load on a two-point support beam
Thanks I got it sorted out with all of your tips :)
@ebaines I meant the deflection :) I thought I'd have to do something like that but wasn't sure how to approach it so thanks for clarifying :)